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Homework Help: Scalar field

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    This is not really a problem but I was going over my lecture notes and I see [itex] \mathscr{H}=\frac{1}{2}\left(\pi^{2} + \vec{\nabla}\phi \cdot \vec{\nabla}\phi + m^{2}\phi^{2}\right) [/itex] and [itex]\frac{\partial\mathscr{H}}{\partial\phi} = -\nabla^{2}\phi + m^{2}\phi [/itex]

    2. Relevant equations

    3. The attempt at a solution

    I would think that [itex]\frac{\partial\mathscr{H}}{\partial\phi} = \nabla^{2}\phi + m^{2}\phi [/itex]. But I don't know where the minus sign is coming from.
  2. jcsd
  3. Sep 29, 2013 #2
    I just found a vector identity [itex] \vec{\nabla}\phi\cdot\vec{\nabla}\phi = \vec{\nabla}\cdot\left(\phi\vec{\nabla}\phi\right) - \phi\vec{\nabla}^{2}\phi [/itex]. I now see how the result follow.

    EDIT: I'm confused again. will the phi derivative of the first term vanish?
  4. Sep 30, 2013 #3
    If by "the first term" you mean the [itex]\pi^2[/itex] term, then yes, the [itex]\phi[/itex] derivative will kill that term; reason being [itex]\phi[/itex] doesn't appear in that term, only its time derivative.

    Edit: Sorry, I understand what you meant by "first term" now. The first term vanishes at infinity so you can ignore it.
    Last edited: Sep 30, 2013
  5. Sep 30, 2013 #4


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    The 3-divergence in the Hamiltonian density (i.e. the first term in the RHS of the equality in post #2) can be discarded, since by integration of full space gives 0, because scalar fields are normally taken from the Schwartz space of test functions.
  6. Oct 2, 2013 #5

    That makes sense. I also realized that it can be integrated by parts using the same reasoning of vanishing at the boundaries. Thanks!
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